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Let $B_2(\mathbb{F}_7)$ be the set of all invertible $2\times2$ upper triangular matrices over $\mathbb{F}_7$. I need to find a composition series for $B_2$. A regular technique I used in previous such questions was to find normal subgroups that their quotient with the main group is of prime order. In here I seem to have a problem achieving this. I have found two normal subgroups: $$ H = \{\begin{pmatrix}x & y \\ 0 & 1\end{pmatrix}\ |\ x\in\mathbb{F}_7^*,\ y\in\mathbb{F}_7\} $$ $$ K = \{\begin{pmatrix}x & a \\ 0 & x\end{pmatrix}\ |\ x\in\mathbb{F}_7^*,\ a\in\mathbb{F}_7\} $$ both of order $6\cdot 7$. The order of $B_2$ is $|B_2| = 6\cdot 6\cdot 7$, so this is still not enough for the same technique. I tried to find a group of order $2$ or $3$ such that the intersection between them is trivial and then take the product of the two, but couldn't find one that is also normal.

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First, what about (with $\;G:=B_2(\Bbb F_7))\;$:

$$N:=\left\{\begin{pmatrix}1&x\\0&1\end{pmatrix}\;|\;x\in\Bbb F_7\right\}\lhd G \;?$$

We clearly have $\;|N|=7\;$ and also $\;N\le H\;$ , so we have $\;1\le N\le H\le G\;$ and also $\;H/N\cong C_6\;$ (why?) . We must refine this step, for example taking

$$\overline T:=\left\langle\begin{pmatrix}2&4\\0&1\end{pmatrix}N\right\rangle=\left\{\begin{pmatrix}2&4\\0&1\end{pmatrix}N\,,\,\,\begin{pmatrix}4&5\\0&1\end{pmatrix}N\,,\,\,\begin{pmatrix}1&0\\0&1\end{pmatrix}N\right\}\cong C_3$$

so we have $\;1\stackrel7\le N\stackrel3\le T\stackrel2\le H\stackrel6\le G\;$ , with $\;T\;$ being such that $\;T/N=\overline T\;$ .

Try now to solve (i.e., refine) the part with $\;G/H\;$ .

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  • $\begingroup$ What about the unit element in $\overline{T}$? it is not equal to any of the matrices, so $\overline{T}$ should be isomorphic to $C_4$, shouldn't it? $\endgroup$ – Joshhh Jan 1 '18 at 9:23
  • $\begingroup$ @Joshhh Thanks for the question: I noticed an ugly mistake I did. Look at the answer again. The elements in $\;H/N\;$ are of the form $\;hN\;$ , with $\;h\in H\;$ ...! Now that part is correct, I believe. $\endgroup$ – DonAntonio Jan 1 '18 at 9:39

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